Question

Following is my code to find middle node of double linked list but it works only for odd number of nodes. i cannot figure out how to take care of case of even number of nodes:

public Object findMiddleNode() {
    DLNode first = getHeadNode();
    DLNode last = getTailNode();
    while(first!=last) {
        first = first.getNext();
        last = last.getPrevious();
    }
    return first.getElement();      
}

---

this is my updated code:

public MyDequeue findMiddleNode() {
    DLNode first = getHeadNode();
    DLNode last = getTailNode();
    MyDequeue m = new MyDequeue();
    while(first!=last && first.getNext()!= last) {
        first = first.getNext();
        last = last.getPrevious();
    }
    if(first == last) {
        m.insertLast(first.getElement());           
        return m;
    }           
    else {
        m.insertLast(first.getElement());
        m.insertLast(last.getElement());
        return m;
    }
}

the class MyDequeue is my implementation of a doubly linked list. the reason I returned MyDequeue is because I already had a static print method that prints the elements of a doubly linked list.

Answer 1

For an even number of nodes, first and last pass each other at the middle. For example, if you have 4 nodes 1..4, first will be 2 when last is 3. At the next iteration, first will be 3, and last will be 2. So you never fail the while loop test (first != last) needed to exit the loop at the proper point. A simple fix is to modify your while test to be

while ((first != last) && (first.getNext() != last)) {.